CS 23/MATH 23
Induction
Proving a statement by mathematical induction is a two step process. In step one, the
basis step, you prove that P(a) is true for a particular integer a. In step two, the
inductive step, you prove that for all integers k a, if P(k)* is true, then P(k + 1)* is
true. *If P(k – 1) is true, then P(k) is true could be substituted here as well.
The inductive step is written formally,
( )
(
)
To prove the inductive step you suppose that P(k) is true, where k is a particular but
arbitrarily chosen integer greater than or equal to a. This supposition is called the
inductive hypothesis.
Next you show that P(k + 1) is true.
1. For each positive integer n, let P(n) be the formula
12 + 22 + 32 + … + n2 = (n(n + 1)(2n + 1))/6
a) Write P(1). Is it true?
b) Write P(k).
c) Write P(k – 1)
d) Write p(k + 1)
2. For each positive integer n, let P(n) be the inequality
n2 < 2n
a) Write P(5). Is it true and why 5?
b) Write P(k).
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CS 23/MATH 23
Induction
c) Write P(k – 1).
d) Write P( k + 1).
3.
Identify the steps in the following proof.
For all integers n 1, 0 + 1 + 2 + 3 + … + n-1 = (n(n-1))/2
Rewrite 1 + 2 + 3 + … + n as a summation
Proof (by mathematical induction)
Show for P(1)
0 = 1(1 - 1)/2
0 = 0/2
0=0
so the formulas is true for n = 1.
Show that for all integers k a,
if the property is true for n = k, then it is true for n = k + 1
Suppose
0 +1 + 2 + 3 + … + k-1 = (k(k-1))/2 for some integer k
We must show that,
1
0 +1 + 2 + 3 + … + (k-1) + k = ((k +1)(k + 1 -1))/2
But,
0 + 1 + 2 + 3 + … + k = 1 + 2 + 3 + … +(k -1) + k //add in the next term which is k
= (k(k-1))/2 + k
= (k(k-1))/2 + (k ) * 2/2 (why?)
= (k(k - 1) + 2k)/2
= (k2 – k + 2k)/2 = (k2 + k)/2
= ((k + 1) k) /2 = ((k + 1) (k + 1 – 1))/2
Substituting n = k + 1 we arrive at (n( n – 1))/2. This is what we needed to prove.
Therefore, by the principle of mathematical induction, the theorem,
For all integers n
1,
0 + 1 + 2 + 3 + … + n-1 = (n(n-1))/2 is true.
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CS 23/MATH 23
Induction
Principle of Direct Inference
P(a) true (premise)
Assume P(k) true for all k 1
Show P(k) P(k + 1) true…..then assumption must be true
P(a)
P(k)
P(k)
P(k + 1)
T
T
T
T
F
F
F
T
T
F
F
T
One more for practice:
4. For each positive integer n, let P(n) be the formula
12 + 22 + 32 + … + n2 = (n(n + 1)(2n + 1))/6
Basis Step:
Inductive Step:
Inductive Hypothesis:
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